,(X)_(n+1))(ngt 1) 取自总体 sim N(mu ,(sigma )^2) , overline (X)=dfrac (1)(n)sum _(i=1)^n(X)_(i) , ^2=dfrac (1)(n-1)sum _(i=1)^n(({X)_(i)-overline (X))}^2 ,则统计量-|||-.=dfrac ({X)_(n+1)-overline (X)}(S)sqrt (dfrac {n)(n+1)} 服从 () .-|||-(A)正态分布 (B)x^2分布 (C)t分布 (D)F分布

,(X)_(n+1))(ngt 1) 取自总体 sim N(mu ,(sigma )^2) . overline (X)=dfrac (1)(n)sum _(i

(B) dfrac (1)(n+1)sum _(i=1)^n(({X)_(i)-overline (X))}^2 .-|||-(C) dfrac (1)(n)s

12.设总体 sim N(mu ,(sigma )^2), μ,σ^2均未知,则 dfrac (1)(n)sum _(i=1)^n(({X)_(i)-overl

设总体 X sim N(mu, sigma^2), mu, sigma^2 均未知，则 (1)/(n) sum_(i=1)^n (X_i - overline(

4.设总体 sim N(mu ,(sigma )^2), x1,x2,···,xn为样本,证明 overline (x)=dfrac (1)(n)sum _(i

,(X)_(n)), 是取自 sim N((mu )_(1)(sigma )^2) 的样本,-|||-X与Y相互独立,若 dfrac (k{(x))^2}(su

4.设X1,.,Xn为正态总体 sim N(mu ,(sigma )^2) 的样本,记 ^2=dfrac (1)(n-1)sum _(i=1)^n(({X)_(

4.样本X1,X2,···Xn来自总体 sim N(0,1) , overline (X)=dfrac (1)(n)sum _(i=1)^n(X)_(i) ,

(B) (hat {sigma )}^2=dfrac (1)(n)sum _(i=1)^n(({X)_(i)-overline (X))}^2.-|||-(C

1.设总体 sim N(M,(sigma )^2) ,μ未知而σ^2已知,X1,X2,···,Xn是X的样本, overline (X)=dfrac (1)(n